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3 years ago

Which acid was most likely used for the titration graph below.

Chemistry

2 answers:

sleet_krkn [62]3 years ago

6 0

Hci acid its high preety cool

Nutka1998 [239]3 years ago

6 0

Answer: HCOOH OR METHANOIC ACID

Explanation:

The titration graph displays the running of alkali or base into an acid. The start of the graph displays a rapid or quick rise in pH but this slows down as a buffer solution contains HCOOH and the unknown base or alkali are produced.

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Choose the term that best fits the following definition: the liquid or gas that carries the sample across the solid support.

iren [92.7K]

The correct answer is option B.

The liquid or gas that carries the sample across the solid support is called mobile phase.

In chromatography, there are two phases: mobile phase and solid phase.

The mobile phase can be either in gas form or liquid form.

While performing the chromatography technique, mobile phase moves over the stationary phase and its components adsorb to the stationary phase and set apart from each other at different rates.

In general, mobile phase refers to the solvent phase that slides over the stationary phase through the chromatography paper.

on the other hand, stationary phase is motionless.

If you need to learn more about gas click here:

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8 0

1 year ago

Marat540 [252]

Answer:

The correct option is A) The light moths will be captured by predators more easily than the dark moths, and the population of dark moths will rise.

Explanation:

As we can see, the colour of the trees do not match with the light moths. Instead, the colour of the trees resembles the dark moths. This adaptation will work best for the dark moths to protect it from its predators. The light moths lack this adaptation and can easily be attacked by the predators. Hence, the population of the dark moth will increase as they are better adapted to live in such an environment.

3 0

3 years ago

Why is the chemical formula 2C NOT an example of a diatomic molecule?

katen-ka-za [31]

Because it represents 2 atoms of carbon not 2 molecules of carbon

8 0

2 years ago

Read 2 more answers

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

5.82 760. Х 425.976 answer should have sig.figs. calculated answer:

Mekhanik [1.2K]

Answer:

2482.4177376

Explanation:

1: 5.82 760 times 425.976

2: You will find the answer that is 2482.4177376

Hope this helps.

7 0

2 years ago